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JEE Advanced 2015Paper 2
Exam
JEE Advanced
Year
2015
Paper
Paper 2
Subject
Mathematics
Chapter
Definite Integration

JEE Advanced 2015 · Paper 2 · Mathematics · Definite Integration

Let F: ℝ → ℝ be a thrice differentiable function. Suppose that F(1)...

Let F:R→RF: \mathbb{R} \to \mathbb{R} be a thrice differentiable function. Suppose that F(1)=0,F(3)=−4F\left(1\right) = 0, F\left(3\right) = - 4 and F′(x)<0F′\left(x\right) < 0 all x∈(12,3).x \in \left(\dfrac{1}{2}, 3\right) . Let f(x)=xF(x)f\left(x\right) = xF ( x ) for all x∈Rx \in \mathbb{R} .
If ∫13x2F′(x)dx=−12\mathop{∫}_{1}^{3}{x}^{2}{F}^{′}\left(x\right)dx = - 12 and ∫13x3F′′(x)dx=40,\mathop{∫}_{1}^{3}{x}^{3}{F}^{′′}\left(x\right)dx = 40, then the correct expression(s) is(are)

Options

  1. A9{f}^{&prime;}\left(3\right) + {f}^{&prime;}\left(1\right) - 32 = 0
  2. B∫13f(x)dx=12\mathop{∫}_{1}^{3}f\left(x\right)dx = 12
  3. C9{f}^{&prime;}\left(3\right) - {f}^{&prime;}\left(1\right) + 32 = 0
  4. D∫13f(x)dx=−12\mathop{∫}_{1}^{3}f\left(x\right)dx = - 12

Correct answer

C. 9{f}^{&prime;}\left(3\right) - {f}^{&prime;}\left(1\right) + 32 = 0

Step-by-step solution

{∫}_{1}^{3}{x}^{2}{F}^{&prime;}\left(x\right)dx = - 12and{∫}_{1}^{3}{x}^{3}{F}^{&prime;&prime;}\left(x\right)dx = 40
{∫}_{1}^{3}{x}^{3}{F}^{&prime;&prime;}\left(x\right) = {x}^{3}{F}^{&prime;}\left(x\right){]}_{1}^{3} - {∫}_{1}^{3}3{x}^{2}F&prime; ( x )
40 = 27{F}^{&prime;}\left(3\right) - {F}^{&prime;}\left(1\right) + 36
4=9(f(3)+4)−f(1)4 = 9\left(f\left(3\right) + 4\right) - f ( 1 )
⇒ 9{f}^{&prime;}\left(3\right) - {f}^{&prime;}\left(1\right) + 32 = 0
Also
{∫}_{1}^{3}{x}^{2}F&prime; ( x ) dx
=x2F(x)]13−∫132xF(x)dx=−12= {x}^{2}F\left(x\right){]}_{1}^{3} - {∫}_{1}^{3}2xF\left(x\right)dx = - 12
9F(3)−F(1)−2∫13xF(x)dx=−129F\left(3\right) - F\left(1\right) - 2{∫}_{1}^{3}xF\left(x\right)dx = - 12
−36−0−2∫13f(x)dx=−12- 36 - 0 - 2{∫}_{1}^{3}f\left(x\right)dx = - 12
−36+12=2∫13f(x)dx- 36 + 12 = 2{∫}_{1}^{3}f ( x ) dx
−12=∫13f(x)dx- 12 = {∫}_{1}^{3}f ( x ) dx
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